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1) In a club there are certain no. of males and females. If
15 females are absent then no. of males will be twice that
of females. If 45 males are absent then female strength
will be 5 times that of males. Find no. of males actually
present.
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Answer Posted / kermit rose
Let the number of males be given the name M.
Let the number of females be given the name F.
If 15 females are absent, then M will be twice that of
present females.
This means that M = 2 * (F - 15)
M = 2 * F - 30.
or 2 * F - M = 30.
Now if have 45 males being absent,
then this gives the equation,
F = 5 * (M - 45)= 5 m - 225
which simplifies to
F = 5 * M - 225
5 * M - F = 225
Pulling the equations together, we get
5 * M - F = 225
-M + 2 * F = 30
Multiply the first equation by 2, and keep the second
equation as is.
10 * M - 2 * F = 450
- M + 2 * F = 30
Add the equations.
9 * M = 480
M = 480/9 is not an integer.
The question is not well stated.
In order to get an integral answer, the question should have been stated as:
1) In a club there are certain no. of males and females. If
15 females are absent then no. of males will be twice that
of females. If 15 females and 45 males are absent then female strength will be 5 times that of males. Find no. of males actually present.
In this form of the question the number of females will be 40 and the number of males will be 50.
(40 - 15) = 25 and 50 = 2 * 25
(50 - 45) * 5 = (40 - 15)
| Is This Answer Correct ? | 1 Yes | 2 No |
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